Найдите нули функции 1) f (x) = |x-5|-1 2) f (x) = |x²+3x|-4 3) f (x) = (x-3) ³ - (x² - 9) 4) f (x) = |x+4|-8 5) f (x) = |x²+11x) - 12 6) f (x) = x³-3x²-4x+12

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Алгебра

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18 мая, 08:01

Найдите нули функции 1) f (x) = |x-5|-1 2) f (x) = |x²+3x|-4 3) f (x) = (x-3) ³ - (x² - 9) 4) f (x) = |x+4|-8 5) f (x) = |x²+11x) - 12 6) f (x) = x³-3x²-4x+12

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Юлий · Answer 1

1) f (x) = |x-5|-1 = 0, |x-5|=1, x1=4, x2 = 6

Ответ: 4; 6.

2) f (x) = |x²+3x|-4 = 0, |x^2 + 3x| = 4

x^2 + 3x - 4 = 0 x^2 + 3x + 4 = 0 нет решений

х1 = - 4; х2 = 1

Ответ: - 4; 1.

3) f (x) = (x-3) ³ - (x² - 9) = 0, (х-3) [x^2 - 6x + 9 - x - 3] = 0

(x-3) (x^2 - 7x+6) = 0, x1 = 3; х2 = 1; х3 = 6.

Ответ: 1; 3; 6.

4) f (x) = |x+4|-8 = 0, |x+4| = 8, x1 = 4, x2 = - 12

Ответ: - 12; 4.

5) f (x) = |x²+11x) - 12 = 0, |x^2 + 11x| = 12

x^2+11x - 12 = 0 x^2+11x+12=0

x1 = - 12, x2 = 1 D = 121-48 = 73

x3 = (кор73 - 11) / 2, х4 = (-кор73 - 11) / 2

6) f (x) = x³-3x²-4x+12 = 0

x^2 (x-3) - 4 (x-3) = (x-3) (x-2) (x+2) = 0

x1=-2; х2=2; х3=3.

Ответ: - 2; 2; 3.